Select all the values of x from this list that make the inequality FOR POINTS!

Which of the following statements regarding the expansion of (x + y)^n are correct?

kompoz [17]

Answer:

A, B and D are true statements.

Step-by-step explanation:

We are given a binomial expansion

$(x+y)^n=^nC_0x^ny^0+^nC_1x^{n-1}y^1+^nC_2x^{n-2}y^2+...........+^nC_nx^0y^n$

$(x+y)^n=x^n+nx^{n-1}y+^nC_2x^{n-2}y^2+...........nxy^{n-1}+y^n$

Now we will check each option

Option A: The coefficients of $x^n$ and $y^n$ both equal 1.

If we see first and last term of the expansion, This statement is true.

Option B: For any term $x^ay^b$ in the expansion, a + b = n.

Let we take 3rd term of expansion $^nC_2x^{n-2}y^2$

Here, a=n-2 and b=2

If we do a+b = n-2+2=n

a+b=n is true statement.

Option C: For any term x^ay^b in the expansion, a - b = n.

Let we take 3rd term of expansion $^nC_2x^{n-2}y^2$

Here, a=n-2 and b=2

If we do a-b = n-2-2=n-4≠n

a-b=n is false statement.

Option D: The coefficients of x^ay^b and x^by^a are equal.

If we take second term from beginning and last of the expansion.

$\text{Coefficient From beginning } nx^{n-1}y=n$

$\text{From last } nxy^{n-1}=n$

This statement true.

8 0

3 years ago

Which graph represents the compound inequality? n < –2 or n ≥ 4

uysha [10]

The second one is the write answer

3 0

2 years ago

Read 2 more answers

Consider a sphere of radius R > 0. Take two antipodal points A and B on the sphere, and another point C on the sphere. Check

aleksklad [387]

Answer:

To this question; we want to show that if two antipodal points on a sphere were A and B, for any random point C on the sphere, AC is perpendicular to BC?

Step-by-step explanation:

I would proceed by imagining the sphere in a three dimensional Cartesian coordinate system. For example, use a sphere of diameter 2 and let it sit at the origin. Then (0, 0, 1) and (0, 0, -1) are the vector locations of the “north and south poles” of the sphere.

Now choose the vector location of any point on the surface of the sphere. It will have a vector location - we’ll call it (x, y, z). Now the vectors from your point to the two poles are (-x, -y, 1-z) and (-x, -y, -1-z).

Now just form the dot product of those two vectors:

(-x, -y, 1-z) . (-x, -y, -1-z) = x^2 +y^2 + (1-z)*(-1-z)

Now the truth of your claim will be embodied in that dot product being zero:

x^2 + y^2 - (1+z)(1-z) = 0

x^2 + y^2 - (1-z^2) = 0

x^2 + y^2 + z^2 = 1

But that last line is just the definition of points on the surface of a sphere of radius 1, so the claim is proven.

Since R>0 and AC is perpendicular to BC, the <ACB is at right angle.

Please, find attached a simple image to show antipodal points (Two points that makes a diameter) and a point C on the sphere.

5 0

3 years ago

Jake went to a fair and spent $5 on admissions, $6.50 on games, and $7.21

asambeis [7]

Answer:

Jake has $11.29 dollars left.

Step-by-step explanation:

To find the answer, we will add up $5 + $6.50 + $7.21 and we get $18.71. Now subtract $30 from 18.71, which is $11.29. Therefore, Jake has $11.29 dollars left.

8 0

3 years ago

Please help me please or i will fail my test please help

stich3 [128]

Answer:

1: B

Step-by-step explanation:

4 0

3 years ago