Answer:
3yz² - 3z² - 5y + 7
Step-by-step explanation:
Sum of two polynomials = –yz² - 3z² – 4y + 4
One of the polynomial = y - 4yz²- 3
Find the other polynomial
The other polynomial = sum of the polynomials - one of the polynomial
= –yz² - 3z² – 4y + 4 - (y - 4yz² - 3)
= –yz² - 3z² – 4y + 4 - y + 4yz² + 3
= -yz² + 4yz² - 3z² - 4y - y + 4 + 3
= 3yz² - 3z² - 5y + 7
A. 0 -2yz?
B. – 4y + 7 01 - 2yz
C. – 3y + 1 0 -5yz² + 3z² – 3y + 1 D. 3yz² - 3z² – 5y + 7
Well, the keyword here is One of the performers insist on being the last.
So, the amount of performers that we can arrange freely is 7 performers.
Different ways we can schedule their appearance is :
7 ! = 7 x 6 x 5 x 4 x 3 x 2 x 1
= 5040
hope this helps
The green arrow thing is a ray. So, question b. is a ray. So the dot would be B and the arrow would be C.
Answer:
30 mph
Step-by-step explanation:
Let d = distance (in miles)
Let t = time (in hours)
Let v = average speed driving <u>to</u> the airport (in mph)
⇒ v + 15 = average speed driving <u>from</u> the airport (in mph)
Using: distance = speed x time

Create two equations for the journey to and from the airport, given that the distance one way is 18 miles:

We are told that the total driving time is 1 hour, so the sum of these expressions equals 1 hour:

Now all we have to do is solve the equation for v:







As v is positive, v = 30 only
So the average speed driving to the airport was 30 mph
(and the average speed driving from the airport was 45 mph)
9/20 is greater
9/20 = 0.45
..............