Quotient: The answer to a division problem
15. If u take 8 and divide it by 12 then u get 3/4. Then multiply 21 by 3/4 (or .75) and it gives you 15.75. Since u can't win three quarters of a game, u round it down to 15
Answer:
since M lies between point A and B , we came to know that,
M(4,6) =(x,y)
A(-2,-1)=(x1 ,y1)
B( _, _ )=(x2,y2)
Now using mid point formula,
x=x1×x2÷2 y=y1+y2÷2
so, point B is (10,13)
Ln ( | Sec (e^x + π) + tan ( e^x +π ) | )
Question 1
f(x) represents the distance of the cannon from the ground
Question 2
When f(x) = 0, there will two values of 'x'. The point on the positive x-axis where the graph crosses represent the point where the cannon hits the ground.
Question 3
Yes, it would. By knowing the spot where the cannon will hit the ground, we can set the net at the spot.
Question 4
f(x) = 0
-0.05 (x² - 26x -120) = 0
x² - 26x - 120 = 0
Question 5
Please refer to the table attached below
Question 6
The value of p and q that gives the correct factors are -30 and 4 since it gives p+q = -26
Question 7
Factorising the equation completely
-0.05 (x² - 26x - 120) = 0
-0.05 (x+4) (x-30) = 0
(x+4) (x-30) = 0
Question 8
Solving the equation
x+4 = 0 and x-30=0
x=-4 and x=30
Question 9
The roots of the equation is x = -4 and x = 30
Question 10
Please refer to the third graph
Question 11
Yes
Question 12
The negative zero means the initial distance of the cannon, where it was fired from
Question 13
The distance between x = -4 and x = 30 is 34 units. If the cannon was fired from the point when x = -4, the cannon will hit the ground again 34 units from the point it was fired from. If Nik put a net 30 units from the firing point, the cannon will fly pass it.