F(x)=x^2+2x+1 & g(x)=3(x+1)^2
now, f(x)+g(x)
=x^2+2x+1+3(x+1)^2
=x^2+2x+1+3(x^2+2x+1)
=x^2+2x+1+3x^2+6x+3
=4x^2+8x+4<===answer(c)
next:
f(x)=x^2-1 & g(x)=x+3
now, f(g(x))=(x+3)^ -1
=x^2+6x+9-1
=x^2+6x+8<====answer(b)
i solve two of ur problems.
now try the 3rd one that is similar to no. 1
and try the last two urself.
Answer:
True
Step-by-step explanation:
A point is zero-dimensional with respect to the covering dimension because every open cover of the space has a refinement consisting of a single open set.
Given:
The x and y axis are tangent to a circle with radius 3 units.
To find:
The standard form of the circle.
Solution:
It is given that the radius of the circle is 3 units and x and y axis are tangent to the circle.
We know that the radius of the circle are perpendicular to the tangent at the point of tangency.
It means center of the circle is 3 units from the y-axis and 3 units from the x-axis. So, the center of the circle is (3,3).
The standard form of a circle is:

Where, (h,k) is the center of the circle and r is the radius of the circle.
Putting
, we get


Therefore, the standard form of the given circle is
.
<h3>
Answer: C) 6</h3>
====================================================
Explanation:
The weird looking E symbol is the greek uppercase letter sigma. It refers to a sum.
It tells us to add up terms in the form (-1)^n*(3n+2) where n is an integer ranging from n = 1 to n = 4.
------------------
If n = 1, then we have
(-1)^n*(3n+2) = (-1)^1*(3*1+2) = -5
Let A = -5 as we'll use it later.
------------------
If n = 2, then
(-1)^n*(3n+2) = (-1)^2*(3*2+2) = 8
Let B = 8 since we'll use this later as well
------------------
If n = 3, then
(-1)^n*(3n+2) = (-1)^3*(3*3+2) = -11
Let C = -11
-------------------
If n = 4, then
(-1)^n*(3n+2) = (-1)^4*(3*4+2) = 14
Let D = 14.
--------------------
We'll add up the values of A,B,C,D to get the final answer
A+B+C+D = -5+8+(-11)+14 = 6
This means that

1 is D
2 is A
3. is B
4. is D
5. is C