Answer:
The number of <u>dimes are 100</u> and number of <u>quarters are 140.</u>
Step-by-step explanation:
Let the number of dimes be 'd' and quarters be 'q'.
Given:
The sum of amount is $45.
The total number of coins are 240.
1 dime = $0.10
∴ 'd' dimes = 
1 quarter = $0.25
∴ 'q' quarters = 
Now, as per question:

Multiplying equation (1) by -0.1 and adding the result to equation (2), we get:

Therefore, the number of dimes are 100 and number of quarters are 140.
Answer:
∠EGF = 65°
Step-by-step explanation:
Since EF = EG the triangle is isosceles and the base angles are equal, that is
∠EGF = ∠GEF
∠EGF =
=
= 65°
Answer:
D
Step-by-step explanation:
Because 90/5=18
Answer:
Step-by-step explanation:
Compounding interest :
Future value of money = Present value * (1+ r)^N
r - interest rate
n - number of period
In our example, Present value = 325.76, FV = 400, r = 2%, and we need to find N
by solving that we can find it that N is equal to 10.3675
Simple interest :
400 - 325.76 = 74.26$ we need to increase
325.76*2% = 6.5152$ each year
74.26 / 6.5152 = 11.3949
as a whole year = 12years
Answer with explanation:
The given differential equation is
y" -y'+y=2 sin 3x------(1)
Let, y'=z
y"=z'

Substituting the value of , y, y' and y" in equation (1)
z'-z+zx=2 sin 3 x
z'+z(x-1)=2 sin 3 x-----------(1)
This is a type of linear differential equation.
Integrating factor

Multiplying both sides of equation (1) by integrating factor and integrating we get

