We can start by solving for 10-5a+(18-b). By plugging in a and b, we get
10 - 10 + (18 - 11)= 7
28 ÷ (a × 2) = 28/4 = 7
20 + b − (4 × a) = 31 - 8 = 23
49 − 43 + (a ÷ a) = 49 - 43 + 1 = 7
30 − 25 + (b − 9) = 5 + 2 = 7
50 ÷ a − (b + 20) = 25 - 31 = -6
So, 28 ÷ (a × 2), 49 − 43 + (a ÷ a), and 30 − 25 + (b − 9) will all have the same answer as 10−5a+(18−b)
Step-by-step explanation:
4(z+2)+4(z+3)
=4z+8+4z+12
=8z+20
by simplifying-
=2z+5
Pull an x from the first two terms
x(x^3 + y^3) + (x^3 + y^3) Now x^3 + y^3 is a common factor.
(x^3 + y^3)*(x + 1) That should be far enough. It can be factored further by factoring (x^3 + y^3) but there is no point because you can't do anything after that. But in case you want to know how x^3 + y^3 factors
(x^3 + y^3) = (x + y)(x^2 - xy + y^2)
Which means you could write original polynomial as
(x + y)(x^2 - xy + y^2)(x + 1)
Part B
You factored the x out of xy^3 so that you would have a common factor (x^3 + y^3) to pull out as a common factor for the whole polynomial.
Evaluate t(1),t(2),t(3),t(4)
t(1)=4.5(1)-8=4.5-8=-3.6
t(2)=4.5(2)-8=9-8=1
t(3)=4.5(3)-8=13.5-8=5.5
t(4)=4.5(4)-8=18-8=10
the first 4 terms are
-3.6,1,5.5, 10
Answer: 27 gallons
Step-by-step explanation:
Since the car can go 35½ miles in 1½ miles gallons, we need to first calculate the number of miles that the car can go in 1 gallon. This will be:
= 35½ / 1½
= 71/2 / 3/2
= 71/2 × 2/3
= 23⅔ miles per gallon.
Therefore, the amount of gallons that the car need to go 639 miles will be:
= 639 / 23⅔
= 639 ÷ 71/3
= 639 × 3/71
= 27
The car will need 27 gallons.
