The answer is 45/4 or 11 1/4
Answer:
Let the smaller number be x
Then the sum is
(6 + 2x) + x = 27
We now find for x
2x + x + 6 = 27
3x + 6 = 27
3x = 27-6
3x = 21
x = 21/3
x = 7
Therefore the smaller number is 7
The larger number is (2×7) + 6 = 20
Answer:
Step-by-step explanation:
![\frac{x+2}{4x^2 + 5x + 1} \ \times \ \frac{4x+1}{x^2-4}\\\\=\frac{x+2}{4x^2 + 4x + x + 1} \ \times \ \frac{4x+1}{x^2-2^2}\\\\=\frac{x+2}{4x(x + 1) + 1( x + 1)} \ \times \ \frac{4x+1}{(x - 2)(x + 2)} \ \ \ \ \ \ \ \ [ \ (a^2 - b^2 = (a-b)(a+b) \ ]\\\\\\=\frac{x+2}{(4x + 1)(x+1)} \ \times \ \frac{4x+1}{(x-2)(x+2)}\\\\=\frac{1}{(x+1)} \ \times \ \frac{1}{(x-2)}\\\\= \frac{1}{(x+1)(x-2)}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%2B2%7D%7B4x%5E2%20%2B%205x%20%2B%201%7D%20%5C%20%5Ctimes%20%5C%20%5Cfrac%7B4x%2B1%7D%7Bx%5E2-4%7D%5C%5C%5C%5C%3D%5Cfrac%7Bx%2B2%7D%7B4x%5E2%20%2B%204x%20%20%2B%20x%20%2B%201%7D%20%5C%20%5Ctimes%20%5C%20%5Cfrac%7B4x%2B1%7D%7Bx%5E2-2%5E2%7D%5C%5C%5C%5C%3D%5Cfrac%7Bx%2B2%7D%7B4x%28x%20%2B%201%29%20%20%2B%201%28%20x%20%2B%201%29%7D%20%5C%20%5Ctimes%20%5C%20%5Cfrac%7B4x%2B1%7D%7B%28x%20-%202%29%28x%20%2B%202%29%7D%20%20%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5B%20%5C%20%28a%5E2%20-%20b%5E2%20%3D%20%28a-b%29%28a%2Bb%29%20%5C%20%5D%5C%5C%5C%5C%5C%5C%3D%5Cfrac%7Bx%2B2%7D%7B%284x%20%2B%201%29%28x%2B1%29%7D%20%5C%20%5Ctimes%20%5C%20%5Cfrac%7B4x%2B1%7D%7B%28x-2%29%28x%2B2%29%7D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B%28x%2B1%29%7D%20%5C%20%5Ctimes%20%5C%20%5Cfrac%7B1%7D%7B%28x-2%29%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B%28x%2B1%29%28x-2%29%7D)
Answer:This can be solved using the strategy used in this post:
It comes down to calculating
(5+3−13)=(73)=7⋅6⋅51⋅2⋅3=35
What we are doing is the following: We create 5 'buckets' of flavors:
−−−|−−−|−−−|−−−|−−−
Now, we want do divide the three scoops over the 5 buckets. Below, a scoop is shown as O. Some possibilities are:
O|O|||O||OOO||etc...
There are 4 borders between the buckets and 3 scoops. Thus, there are (4+33)=35 possibilities
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
Under a reflection in the y- axis
a point (x, y ) → (- x, y ) , then
A(- 1, 3 ) → A'(1, 3 ) → C
