So, first we multiply the fraction by using the formula a/b times c/d= a times c/b times d
=(y^2-16) times 5y/2y(y-4)
Now, we cancel the common factor y
=(y^2-16) times 5/2(y-4)
Now, we factor 5(y^2-16)
We factor (y^2-16) first
y^2-16
Rewrite 16 as 4^2
y^2-4^2
Now, apply the formula x^2-y^2=(x+y)(x-y)
=y^2-4^2=(y+4)(y-4)
=5(y+4)(y-4)
=5(y+4)(y-4)/2(y-4)
Cancel the common factor y-4
=5(y+4)/2
Answer: 5(y+4)/2
Answer:
Yes, you are correct, if it is 37 minutes and you are supposed to round, you round up.
Step-by-step explanation:
Answer:
refer to explanation
Step-by-step explanation:
the lengths of the base and height are the factors of 20
so they can be 5×4, 10×2, 1×20
Answer:
Option D. (x + 4)(x + 1)
Step-by-step explanation:
From the question given above, the following data were obtained:
C = (6x + 2) L
D = (3x² + 6x + 9) L
Also, we were told that half of container C is full and one third of container D is full. Thus the volume of liquid in each container can be obtained as follow:
Volume in C = ½C
Volume in C = ½(6x + 2)
Volume in C = (3x + 1) L
Volume in D = ⅓D
Volume in D = ⅓(3x² + 6x + 9)
Volume in D = (x² + 2x + 3) L
Finally, we shall determine the total volume of liquid in the two containers. This can be obtained as follow:
Volume in C = (3x + 1) L
Volume in D = (x² + 2x + 3) L
Total volume =?
Total volume = Volume in C + Volume in D
Total volume = (3x + 1) + (x² + 2x + 3)
= 3x + 1 + x² + 2x + 3
= x² + 5x + 4
Factorise
x² + 5x + 4
x² + x + 4x + 4
x(x + 1) + 4(x + 1)
(x + 4)(x + 1)
Thus, the total volume of liquid in the two containers is (x + 4)(x + 1) L.
