Multiply the numerators: 7×8=56 then the denominators 10×21=210. The lowest multiples for 56 and 210 is 14. So divide 56/14=4 and 210/14=15. Your answer is 4/15.
Hope this helps (:
Given:
Markers are sold in packages of 12 and pens are sold in packages of 8.
To find:
Least amount of markers and pens Rick needs to buy to have an equal number of markers and pens
Solution:
To find the least amount of markers and pens, we need to find the LCM of 12 and 8.
Prime factors of 8 and 12 are
To find the LCM, multiply all factors but the common factors are included only once.
Therefore, the least amount of markers and pens Rick needs to buy to have an equal number of markers and pens is 24.
It would be 46, have a nice day
Answer: Mode = 12
Explanation: The mode is the most frequent value, so it is the value that shows up the most. In this case, that would be the value "12" since it shows up three times (while the other values only show up once).
Answer:
Parking lot is 60m wide
Step-by-step explanation:
In this question. We are asked to calculate the width of the parking lot. We proceed as follows:
We know that the total length of the field and the parking lot is 300m while the total width is 200m. We should remember that the field is rectangular in nature.
Since the parking lot is x meters, the length of the field would simply be (300-2x)m while the width of the field would be (200-x). The width would also have been 200-2x but we must remember that the field is only covered on 3 sides.
Area of the field is 30,000. This means if we multiply both, we get 30,000.
Mathematically;
(300-2x)(200-x) = 30,000
60,000-300x-400x+ 2x^2 = 30,000
2x^2-700x-30,000= 0
We can see we now have a quadratic equation we need to solve.
2x^2+500x-1200x-30,000= 0
2x(x+250) -120(x+250) = 0
(2x-120)(x+250) = 0
2x-120= 0 or x+250=0
2x = 120or x = -250
x = 120/2 or x= -250
x = 60 or x=-250
We discard x = -100 as x is length and cannot be negative
Hence, x = 60m
