Let
x = first integer
y = second integer
z = third integer
First equation: x + y + z = 194
Second equation: x + y = z + 80
Third equation: z = x - 45
Let's find the values of x, y and z.
Substitute 3rd eq to 1st eq:
x + y + x - 45 = 194
2x + y = 45 + 194
y = -2x + 239
Plug in both we have solved for y and the 3rd eq to the 2nd eq to find x
x + (-2x + 239) = (x - 45) + 80
x - 2x - x = -45 + 80 - 239
-2x = -204
x = -204/-2
x = 102
Solving for y,
y = -2(102) + 239
y = 35
Solving for z,
z = 102 - 45
z = 57
Hi!
C. v + 15
Barry read for 15 MORE minutes than Robert.
Answer:
the answer would be 4
Step-by-step explanation:
10 - 4 = 6 divided by 1.25 = 4
hope this helps!
Step-by-step explanation:
We need to use the binomial theorem/Pascal's triangle here.
(a+b)^5 = (5 choose 0)a^5 + (5 choose 1)a^4*b + (5 choose 2)a^3*b^2 + (5 choose 3)a^2*b^3 + (5 choose 4)a*b^4 + (5 choose 5)b^5.
5 choose 0 = 1
5 choose 1 = 5
5 choose 2 = 10
5 choose 3 = 10
5 choose 4 = 5
5 choose 5 = 1
And 1, 5, 10, 10, 5, 1, is the (5+1) = 6th row of pascal's triangle.
Therefore we get
g^5 + 5g^4*2 + 10g^3*2^2 + 10g^2*2^3 + 5g*2^4 + 2^5
which is
g^4 + 10g^4 + 40g^3 + 80g^2 + 80g + 32
Or, you could do the slow way, by just doing (g+2)(g+2)(g+2)(g+2)(g+2)