Y = x + 2 is closest to what I believe is the solution.
If the club begins with 2 members, then +2 MUST appear in the formula.
If the club adds 2 members every week, then the total number of members would be
f(x) = y = 2 + 2(x-1), where x is the number of weeks.
Test this: If x = 1, then y = 2+2(1-1) = 2 + 0 = 2. This is correct. The club began with 2 members in week 1
<span>A
gardener has 27 pansies and 36 daises. He planted equal parts of each flowers.
Now we need to find the value of flowers did the gardener planted.
If you’re looking for the greatest number of flowers planted in a row.
We can have 27 pansies and 27 daises.
However, if we want to plant all flowers we can have:
3 rows of pansies with 9 flowers = (3 x 9) = 27
4 rows of daises with 9 flowers = (4 x 9) = 36</span>
In 2005 he average weight is 4 ounces, they have gotten bigger since, clearly, but that is the average weight.
Does this answer your question?<span />
Answer:
Below.
Step-by-step explanation:
f) (a + b)^3 - 4(a + b)^2
The (a+ b)^2 can be taken out to give:
= (a + b)^2(a + b - 4)
= (a + b)(a + b)(a + b - 4).
g) 3x(x - y) - 6(-x + y)
= 3x( x - y) + 6(x - y)
= (3x + 6)(x - y)
= 3(x + 2)(x - y).
h) (6a - 5b)(c - d) + (3a + 4b)(d - c)
= (6a - 5b)(c - d) + (-3a - 4b)(c - d)
= -(c - d)(6a - 5b)(3a + 4b).
i) -3d(-9a - 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b).
= (3d + 2c)(9a + 2b).
j) a^2b^3(2a + 1) - 6ab^2(-1 - 2a)
= a^2b^3(2a + 1) + 6ab^2(2a + 1)
= (2a + 1)( a^2b^3 + 6ab^2)
The GCF of a^2b^3 and 6ab^2 is ab^2, so we have:
(2a + 1)ab^2(ab + 6)
= ab^2(ab + 6)(2a + 1).
You will need to drain 1016.308 gallons an hour
