ensure all fractions have the same denominator
we require -
= - ![\frac{4}{16}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B16%7D)
thus -
-
-
take out a common factor ![\frac{1}{16}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B16%7D)
=
(- 7 - 4 - 5 )
=
× - 16 = - 1
Answer:
The area of the circle in terms of π is:
Step-by-step explanation:
To find the area of that circle, you can use the equation:
- Area of a circle =
![\pi * \frac{D^{2}}{4}](https://tex.z-dn.net/?f=%5Cpi%20%2A%20%5Cfrac%7BD%5E%7B2%7D%7D%7B4%7D)
Where:
Now, we can replace the given measurement in the equation:
- Area of a circle =
![\pi * \frac{(\frac{3}{4}in) ^{2}}{4}](https://tex.z-dn.net/?f=%5Cpi%20%2A%20%5Cfrac%7B%28%5Cfrac%7B3%7D%7B4%7Din%29%20%5E%7B2%7D%7D%7B4%7D)
- Area of a circle =
![\pi * \frac{\frac{9}{16}in ^{2}}{4}](https://tex.z-dn.net/?f=%5Cpi%20%2A%20%5Cfrac%7B%5Cfrac%7B9%7D%7B16%7Din%20%5E%7B2%7D%7D%7B4%7D)
- Area of a circle =
![\pi * \frac{9}{64} in^{2}](https://tex.z-dn.net/?f=%5Cpi%20%2A%20%5Cfrac%7B9%7D%7B64%7D%20in%5E%7B2%7D)
That result is the same that to write:
- Area of a circle =
![\frac{9}{64}\pi in^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B64%7D%5Cpi%20%20in%5E%7B2%7D)
By this reason, <u><em>the result is 9/64 π in^2, giving the result in terms of π</em></u>.
Answer: We should dispense 60 tablets.
Step-by-step explanation:
Given : A patient is to receive 2 tablets po ACHS for 30 days.
i.e. Dose for each day = 2 tablets
Number of days = 30
If a patient takes 2 tablets each day , then the number of tablets he require for 30 days will be :-
[Multiply 2 and 30]
Therefore, the number of tablets we should dispense = 60
To solve this problem, we just need to divide his total payment by the number of monthly payments he made:
14256 ÷ 48 = 297
Tom's monthly payment was $297.
1) Look for common factors. You see that y^2 is a factor of every term so you can remove it to get
... = (y^2)(3x^2 -2x -8)
The quadratic in x can be factored by your favorite method. There is one called by various names that has you look for factors of (3)(-8) that add to (-2). When the quadratic is written as ax^2+bx+c, you're looking for factors of the product "ac" that add to "b". Of course, you know that
... -24 = -24*1 = -12*2 = -8*3 = -6*4
the last factor pair shown here has a sum of -2, so our factorization is
... = (y^2)(3x -6)(3x +4)/3 . . . . . the "a" coefficient is repeated in each factor (at first), then divided out
... = (y^2) (x -2) (3x +4)
2) You recognize this expression to be of the form
... (x +a)^2 = x^2 +2ax + a^2
where a=5. As a result, you know the factorization is
... = (x +5)^2
3) You recognize this expression to be the difference of squares, so you know the factorization is
... a^2 - b^2 = (a -b)(a +b)
where a=x and b=6. As a result, you know the factorizatin is
... = (x -6) (x +6)