Answer:
(x+3)(x+3)
Step-by-step explanation:
1) this polynomial is in the form of ax^2+bx+c
c=9
b=6
a=1
what*what=c(9)
and what+what=b(6)
3 is the answer to both questions:
now substitute "3x" in place of "6"
x^2+3x+3x+9
split in 2 groups:
group a) (x^2+3x)
gcf of group a=x
x(x+3)
group b) (3x+9)
gcf of group b=3
3(x+3)
combine the numbers OUTSIDE the parantheses
so
(x+3)(x+3)
Answer:
Step-by-step explanation:
We are asked to find the integral:
So, let's apply the vertices given in the question.
![\int\limits^1_0\int\limits_{2x}^{3-x}4xy\:dydx=\int\limits^1_0 2xy^2|_{2x}^{3-x}\:dx=\int\limits^1_0 2x[(3-x)^2-(2x)^2]\:dx=\\\\=\int\limits^1_0 2x(9-6x-3x^2)\:dx=6\int\limits^1_0 (3x-2x^2-x^3)\:dx=\\\\=6(\frac{3}{2}x^2-\frac{2}{3}x^3-\frac{1}{4}x^4)|^1_0=6(\frac{3}{2}-\frac{2}{3}-\frac{1}{4})=\frac{7}{2}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E1_0%5Cint%5Climits_%7B2x%7D%5E%7B3-x%7D4xy%5C%3Adydx%3D%5Cint%5Climits%5E1_0%202xy%5E2%7C_%7B2x%7D%5E%7B3-x%7D%5C%3Adx%3D%5Cint%5Climits%5E1_0%202x%5B%283-x%29%5E2-%282x%29%5E2%5D%5C%3Adx%3D%5C%5C%5C%5C%3D%5Cint%5Climits%5E1_0%202x%289-6x-3x%5E2%29%5C%3Adx%3D6%5Cint%5Climits%5E1_0%20%283x-2x%5E2-x%5E3%29%5C%3Adx%3D%5C%5C%5C%5C%3D6%28%5Cfrac%7B3%7D%7B2%7Dx%5E2-%5Cfrac%7B2%7D%7B3%7Dx%5E3-%5Cfrac%7B1%7D%7B4%7Dx%5E4%29%7C%5E1_0%3D6%28%5Cfrac%7B3%7D%7B2%7D-%5Cfrac%7B2%7D%7B3%7D-%5Cfrac%7B1%7D%7B4%7D%29%3D%5Cfrac%7B7%7D%7B2%7D)
Answer:
12 and 15
Step-by-step explanation:
x + y = 27
x +3 = y
x = y - 3
y - 3 + y = 27
2y = 24
y = 12
Answer:
-3/2, 1.25, |-9/4|, 2.5, 121
Step-by-step explanation:
Convert all numbers to a decimal
-3/2 = -1.5
|-9/4| = 2.25
2.5 = 2.5
121 = 121.0
1.25 = 1.25
Now put them in order from least to greatest.
-3/2, 1.25, |-9/4|, 2.5, 121
