Y^4 + 4xy^3 (y') + 2xy + x^2 (y') = 1 + 3 (y')
(y')(4xy^3 + x^2 - 3) = 1 -y^4 -2xy
I'll let you do the next line
First, we'll set up two equations. One for the amount of each coin and another for the value of the coins.
N will represent nickels
D will represent dimes
N + D = 30
---The problem tells us that there are 30 total coins
0.05N + 0.10D = 2.95
---Nickels are worth 5 cents and dimes are worth 10 cents, and the total value of the coins is 2.95
Now that we have our equations, we need to solve for one of the variables in the first equation. I will solve for N.
N + D = 30
N = 30 - D
Then, we take that equation and substitute our new value for N into the second equation (value) and solve for D.
0.05(30 - D) + 0.10D = 2.95
1.5 - 0.05D + 0.10D = 2.95
1.5 + 0.05D = 2.95
0.05D = 1.45
D = 29
Now that we know how many dimes there are, we can plug that value into our equation for N and solve for N.
N = 30 - D
N = 30 - 29
N = 1
Therefore, there are 29 dimes and 1 nickel.
Hope this helps!
Answer:
middle term = 7x
Step-by-step explanation:
in order to find the middle term, we will expand the bracket on the left hand side and compare the expressions on both sides.
(x + 2) (3x + 1) = 3x² + ___ + 2
let the unknown expression = y
(x + 2) (3x + 1) = 3x² + y + 2
Expanding the bracket on the left hand side by multiplication:
x + 2( 3x + 1) = (x × 3x) + (x × 1) + (2 × 3x) + ( 2 × 1)
= 3x² + x + 6x + 2
= 3x² + 7x + 2
Comparing with the expression on the right hand side:
3x² + 7x + 2 = 3x² + y + 2
∴ y = 7x
middle term = 7x