Answer:
x=15
Step-by-step explanation:
Step 1: Multiply both sides by 2.
Step 2: Simplify both sides of the equation.
6x+18=108
Step 3: Subtract 18 from both sides.
6x+18−18=108−18
6x=90
Step 4: Divide both sides by 6.
x=15
Answer:
Part A:
(1) x + y = 95
(2) x = y + 25
Part B:
The number of minutes Eric spends playing volleyball each day is 35 minutes
Part C:
It is not possible for Eric to have spent exactly 35 minutes playing basketball
Step-by-step explanation:
The total time Eric plays basketball and volleyball = 95 minutes
The time duration Eric plays basket ball = x
The time duration Eric plays volleyball = y
Part A:
The pair of relationships between the number of minutes Eric plays basketball (x) and the number of minutes he plays volleyball (y) are;
(1) x + y = 95
(2) x = y + 25
Part B:
By substituting the value of x in equation (2) into equation (1), we have;
x + y = (y + 25) + y = 95
2·y + 25 = 95
2·y = 95 - 25 = 70
y = 70/2 = 35 minutes
Therefore, Eric spends 35 minutes playing volleyball every day
Part C:
It is not possible for Eric to have spent only 35 minutes playing basketball because, given that he plays basketball for 25 minutes longer than he plays volley, the number of minutes he spends playing volleyball will then be given as follows;
x = y + 25
35 = y + 25
y = 35 - 25 = 10 minutes
The total time = x + y = 10 + 35 = 45 minutes ≠ 95 minutes.
Answer:
2201.8348 ; 3 ; x / (1 + 0.01)
Step-by-step explanation:
1)
Final amount (A) = 2400 ; rate (r) = 6% = 0.06, time, t = 1.5 years
Sum = principal = p
Using the relation :
A = p(1 + rt)
2400 = p(1 + 0.06(1.5))
2400 = p(1 + 0.09)
2400 = p(1.09)
p = 2400 / 1.09
p = 2201.8348
2.)
12000 amount to 15600 at 10% simple interest
A = p(1 + rt)
15600 = 12000(1 + 0.1t)
15600 = 12000 + 1200t
15600 - 12000 = 1200t
3600 = 1200t
t = 3600 / 1200
t = 3 years
3.)
A = p(1 + rt)
x = p(1 + x/100 * 1/x)
x = p(1 + x /100x)
x = p(1 + 1 / 100)
x = p(1 + 0.01)
x = p(1.01)
x / 1.01 = p
x / (1 + 0.01)
Answer:
she received 364.80 euros
Step-by-step explanation:
The action of being active
