Form the union for the following sets. M = { } N = {6, 7, 8, 9, 10} M ∪ N =

Find the mean, median, mode for the data set below. 32, 20, 45, 20, 14, 7, 20

Butoxors [25]

Answer:

mean-22

mode-20

median-20

7 0

3 years ago

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Given the quadratic equation below, determine which situation it could represent? x2 -4x = 60

yarga [219]

Answer:

The solution is x = 10 and x = -6. This quadratic could be represented in factored form as (x - 10)(x + 6) or on a graph with x-intercepts at (10,0) and (-6,0).

Step-by-step explanation:

To solve the quadratic, write the quadratic in standard form and factor the equation.

x² - 4x = 60

x² - 4x - 60 = 0

(x - 10)(x + 6) = 0

x = 10 and x = -6

4 0

3 years ago

Help! What is the average rate of change of f(x)=x^2+3x+6 over the interval -3 less-than-or-equal-to x less-than-or-equal-to 3?

Makovka662 [10]

Answer:

The average rate of change of the function over the interval is 5.

Step-by-step explanation:

Average rate of change of a function:

The average rate of change of a function f(x) over an interval [a,b] is given by:

$A = \frac{f(b)-f(a)}{b-a}$

Interval -3 less-than-or-equal-to x less-than-or-equal-to 3

This means that $a = -3, b = 3$

$f(x) = x^2 + 3x + 6$

So

$f(b) = f(3) = (3)^2 + 3(3) + 6 = 9 + 9 + 6 = 24$

$f(a) = f(-3) = (-3)^2 + 3(-3) + 6 = 9 - 9 + 6 = 6$

Average rate of change

$A = \frac{f(b)-f(a)}{b-a} = \frac{24+6}{3-(-3)} = \frac{30}{6} = 5$

The average rate of change of the function over the interval is 5.

8 0

3 years ago

Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

4 0

2 years ago

I bet you want Brainlest!! Click HERE!!! Please help me!!! BRaNiElsT!!!!! Click!!!

Ilia_Sergeevich [38]

What? What's your question?

3 0

3 years ago

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