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iVinArrow [24]
2 years ago
9

Which of the following describes the roots of the polynomial function f(x)=(x-3)^4(x+6)^2

Mathematics
1 answer:
yulyashka [42]2 years ago
3 0

Answer:y=3x+4

Step-by-step explanation:

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What is the solution set of {x | x < -3} ∩ {x | x > 5}
lozanna [386]
Any x in this set will be real numbers that are both less than -3 and greater than 5. But that's not possible, so this set is empty.
7 0
3 years ago
If T is the midpoint of segment RS then, Segment RT is congruent to segment TS..<br> Help please!!!!
omeli [17]

Answer:

Definition of Midpoint

Step-by-step explanation:

Since T is the midpoint then it is equadistant from R to T and T to S

3 0
3 years ago
Parametros pára describir el movimiento
VikaD [51]

Answer:

Step-by-step explanation:

Parameters to describe the movement

6 0
2 years ago
X/-3-41=-11 show work please
Brilliant_brown [7]

The correct answer is x = -90.

To find this, solve using the order of operations. See the example below.

x/-3 - 41 = -11 ----> Add 41 to both sides

x/-3 = 30 ----> Multiply each side by -3

x = -90

4 0
3 years ago
Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.
NISA [10]

Answer:

\frac{\sqrt{\pi}}{4}

Step-by-step explanation:

You are going to integrate the following function:

g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}  (1)

furthermore, you know that:

\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}

lets call to this integral, the integral Io.

for a general form of I you have In:

I_n=\int_0^{\infty}x^ne^{-ax^2}dx

furthermore you use the fact that:

I_n=-\frac{\partial I_{n-2}}{\partial a}

by using this last expression in an iterative way you obtain the following:

\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}} (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}

5 0
2 years ago
Read 2 more answers
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