57 but it is an estimate because of the 4 so your answer is 57.4
Answer: Amira added 5 candy bars to the p pile of candy bars there were no more than 18 candy bars in the pile.
Step-by-step explanation: hope this helps
Answer:
The equation that describes the amount of mass left after a time t of a radioactive isotope is the following:
where
is the mass of the sample at t = 0
is the half-life of the sample
For the element X in this problem,
We want to find the time t at which
So we need to re-arrange the equation making t the subject:
Step-by-step explanation:
Answer: Rs 2,184
Explanation:
1) The statement is incomplete. The complete statement contains the information of the dimensions of both bigger and smaller cardboard boxes.
2) The dimesions of bigger cardboard boxes are 25cm * 20 cm * 5 cm
3) The dimensions of smaller cardboard boxes are 15 cm * 12 cm * 5cm
4) For bigger cardboard boxes:
length, l = 25 cm
width, w = 20 cm
height, h = 5 cm
surface of each bigger carboard box = 2 [ l*w + l*h + w*h] = 2 [25*20 + 25*5 + 20*5] cm^2 = 1450 cm^2
total surface of 250 bigger cardboard boxes = 250 * 1450 cm^2 = 3625,500 cm^2
5% of the total surface area extra = 362,500cm^2 * 5 / 100 = 18,125 cm^2
Total area for bigger cardboard boxes= 362,500 cm^2 + 18,125 cm^2 = 380,625 cm^2
5) Smaller cardboard boxes
length, l = 15 cm
width, w = 12 cm
height, h = 5 cm
surface of each smaller cardboard box = 2 [l*w + l*h + w*h] = 2 [ 15*12 + 15 * 5 + 12 * 5] cm^2 = 630 cm^2
total surface of 250 smaller cardboard boxes = 250 * 630 cm^2 = 157,500 cm^2
5 % extra = 157,500 cm^2 * 5 / 100 = 7,875 cm^2
total area for smaller cardboard boxes = 157,500 cm^2 + 7,875 cm^2 = 165,375 cm^2
6) total area of cardboard required = 380,625 cm^2 + 165,375 cm^2 = 546,000 cm^2
7) Cost of cardboard required
unit cost per area * total area = (Rs 4 / 1000cm^2) * 546,000 cm^2 = Rs 2,184.
Answer: Rs 2,184
Answer:
The answer to the question is 10a√2a(b + 2)
Step-by-step explanation:
To know the student's error, let us calculate the perimeter from the the start.
Dimension of the rectangle is given below:
√(50a³b²) and √(200a³)
Perimeter of a rectangle = 2(L + B) = 2L + 2B
2√(50a³b²) + 2√(200a³)
2 x 5ab√2a + 2 x 10a√2a
10ab√2a + 20a√2a
Factorise
10a√2a (b + 2)
Therefore, the student made an error by adding 10ab√2a and 20a√2a together.
Note: we can not add 10ab√2a and 20a√2a together since b is missing in one of the expression.
Therefore, the answer to the question is 10a√2a(b + 2).
