1.
1.2 x^4 ( 4 x² + 3 x + 1 ) = 4.8 x^6 + 3.6 x^5 + 1.2 x^4
2.
( 4 x - 3 ) ( 2 x² - 7 x + 1 ) = 8 x³ - 28 x² + 4 x - 6 x² + 21 x - 3 =
= 8 x³ - 34 x² + 25 x - 3
3.
( x² + 4 x - 3 ) ( 2 x² + x + 6 ) =
= 2 x^4 + x³ + 6 x² + 8 x³ + 4 x² + 24 x - 6 x² - 3 x - 18 =
= 2 x^4 + 9 x³ + 4 x² + 21 x - 18
Step-by-step explanation:
5x^2+2x=1
5x^2+2x-1=0
x=-b+/-√b^2-4ac÷2a
x=-2+/-√-2^2-(4×5×-1)÷2×5
x=-2+/+√4--20÷10
x=-2+/-√24÷10
x=-2+√24÷10 or -2-√24÷10
x=-2+4.9÷10 or -2-4.9÷10
x=2.9÷10 or -6.9÷10
x=0.29 or -0.69
Answer:
3a-b/(x-b)=1/(x+1)Doing criss cross multiplication
3a(x+1)-b(x+1)=x-b
3ax+3a-bx-b=x-b
3ax-bx-x=b-3a-b
x(3a-b-1)=-3a
x=3a/(3a-b-1)
Answer:
Step-by-step explanation:
I can't get to act correctly either.
The answer is 1 day and 7 items.
Ariana
y = 5x + 2
Julie
y = 7x
The ys have to be the same so equate the right hand side.
7x = 5x + 2 Subtract 5x from both sides.
7x - 5x = 5x - 5x + 2
2x = 2 Divide by 2
2x/2 = 2/2
x = 1
1 day
7 items
Answer:
b) Geometric
Step-by-step explanation:
given that the a quarter back completes 44% of his passes. We want to observe this quarterback during one game to see how many pass attempts he makes before completing one pass.
This is a geometric distribution because we are considering the random variable as number of attempts he makes before passing first time.
It is not binomial since number of trials is not fixed
It is a geometric distirbution.
Option b is right
Recall that a geometric distribution is the number of trials before you get first success.
