<span>With the bag having 4/5 lb currently in it, this can be translated to 16/20 lb. Maria wants to scoop out sand until the bag has 1/2 lb, which is 10/20 lb. The scoop that Maria uses is 1/20 lb, so using the equation of 16-10 = 6, Maria can get 6 scoops of sand before the bag contains 1/2 lb.</span>
Answer:
A, C, E
Step-by-step explanation:
From the table you can see that the water depth cahnges
for every
of snow (option B is false).
This means that the function modelling this situation is linear function (option A is true and option D is false). Let the equation of this function be 
Then
Subtract these two equations:
Hence,
The equation of the straight line (the graph of linear function) is 
(option E is true) This line passes through the point (0,0), because its coordinates satisfy the equation (option C is true).
The solution
m = s + 5
<span>m - 4 = (3s) - 3 </span>
<span>(s + 5) - 4 = 3s - 3 </span>
<span>s + 1 = 3s - 3 </span>
<span>s = 3s - 4 </span>
<span>-2s = -4 </span>
<span>s = 2 </span>
<span>m = 7 </span>
<span>Check this: </span>
<span>m - 4 = 3s - 3 </span>
<span>7 - 4 = 3(2) - 3 </span>
<span>3 = 6 - 3 </span>
<span>3 = 3 </span>
<span>So Sandy is 2 and Megan is 7.</span>
Answer:
1. 15
2. 8
Step-by-step explanation:
The two sequence are geometric progression GP, because they follow a constant multiple (common ratio)
The nth term of a GP is;
Tn = ar^(n-1)
Where;
a = first term
r = common ratio
For the first sequence;
The common ratio r is
r = T3/T2 = 540/90 = 6
r = 6
T2 = ar^(2-1) = ar
T2 = 90 = ar
Substituting the values of r;
90 = a × 6
a = 90/6
a = 15
First term = 15
2. The sam method applies here.
Common ratio r = T3/T2 = 128/32 = 4
r = 4
T2 = ar^(2-1) = ar
T2 = 32 = ar
Substituting the values of r;
32 = a × 4
a = 32/4
a = 8
First term = 8
