Answer:
Ihorangi uses 7.65 litres of petrol to and from work
Step-by-step explanation:
If the car Ihorangi drives consumes 7.5 L/100 km
Then for 1 km journey, the car will consume 7.5/100 L
= 0.075 L
If Ihorangi travels 51 km to work and 51 km from work
Therefore Ihorangi travel (51 + 51) km daily to and from work
= 102 km daily
The fuel he consumes daily is 102*0.075 L
= 7.65 L
-3+8.
We know that the temperature /was/ -3, so it was 3 under zero. It has /risen/ by 8, meaning we have added 8 degrees to whatever the starting temperature was.
We can find out how much the temperature is right now: it was -3, went through -2, -1, 0, 1, 2, 3, 4, and now its at 5 degrees. Thats risen by 8 degrees, right? And -3+8 is the same as 8-3 (commutative property of addition, since we basically did 8+(-3)) which equals 5, so we know that we got the correct answer. :)
Answer:
-2, 3, -0.5 + 0.866i, -0.5 - 0.866i.
Step-by-step explanation:
As the last term is -6 , +/- 2 , +/- 3 are possible zeroes.
Try 2:-
(2)^4 - 6(2)^2 - 7(2) - 6 = -28 so 2 is not a zero.
3:-
(3)^4 - 6(3)^2 - 7(3) - 6 = 0 so 3 is a zero.
(-2)^4 - 6(-2)^2 - 7(-2) - 6 = 0 so -2 is also a zero.
Divide the function by (x +2)(x - 3), that is x^2 - x - 6
gives x^2 + x + 1
x^2 + x + 1
So we have x^2 + x + 1 = 0
x = [-1 +/- √(1^1 - 4*1*1)] / 2
= -1 + √(-3) / 2 , - 1 - √(-3) / 2.
= -0.5 + 0.866i, -0.5 - 0.866i
Answer:
okayyyyy.... then
Step-by-step explanation:
Step-by-step explanation:
solve for surface area
A=6a^2
a=edge
