What is the mean absolute deviation for the data set?

Find the x and y intercept of the line below 2x + 5y = 10

Dmitriy789 [7]

Answer:

x - intercept = 5

y- intercept = 2

Step-by-step explanation:

2x + 5y = 10

Write the equation in $\frac{x}{a}+\frac{y}{b}=1$ form

So, divide the equation by 10

$\frac{2x}{10}+\frac{5y}{10}=\frac{10}{10}\\\\\frac{x}{5}+\frac{y}{2}=1$

x - intercept = 5

y- intercept = 2

5 0

3 years ago

Read 2 more answers

How many solutions does 17.75x + 24 = 18.95x + 18 have?

iren2701 [21]

Answer: Simplifying

17.75x + 24 = 18.95x + 18

Reorder the terms:

24 + 17.75x = 18.95x + 18

Reorder the terms:

24 + 17.75x = 18 + 18.95x

Solving

24 + 17.75x = 18 + 18.95x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-18.95x' to each side of the equation.

24 + 17.75x + -18.95x = 18 + 18.95x + -18.95x

Combine like terms: 17.75x + -18.95x = -1.2x

24 + -1.2x = 18 + 18.95x + -18.95x

Combine like terms: 18.95x + -18.95x = 0.00

24 + -1.2x = 18 + 0.00

24 + -1.2x = 18

Add '-24' to each side of the equation.

24 + -24 + -1.2x = 18 + -24

Combine like terms: 24 + -24 = 0

0 + -1.2x = 18 + -24

-1.2x = 18 + -24

Combine like terms: 18 + -24 = -6

-1.2x = -6

Divide each side by '-1.2'.

x = 5

Simplifying

x = 5

Step-by-step explanation:

3 0

3 years ago

2x + x = please help

Troyanec [42]

Well my man, if it does not give you a value for x.... your answer would be 2x

5 0

3 years ago

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True or false: an cubic function may have 3 irrational zeros

jolli1 [7]

<h3>Answer: True</h3>

The key word here is "may" meaning that we could easily have 3 rational roots as well. An example of a cubic having 3 irrational roots would be

(x-1)(x-2)(x-3) = x³ - 6x² + 11x - 6

This has the rational roots x = 1, x = 2, x = 3.

However, we could easily replace 1,2,3 with any irrational numbers we want. So that's why the statement "a cubic has three irrational roots" is sometimes true.

In some cases, a cubic may only have 1 real root and the other 2 roots are imaginary.

3 0

3 years ago

In this problem, you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+4y=12te^(−2t)−(8t+12) with

Zarrin [17]

First check the characteristic solution:

y'' + 4y' + 4y = 0

has characteristic equation

r ² + 4r + 4 = (r + 2)² = 0

with a double root at r = -2, so the characteristic solution is

$y_c = C_1e^{-2t} + C_2te^{-2t}$

For the particular solution corresponding to $12te^{-2t}$ , we might first try the ansatz

$y_p = (At+B)e^{-2t}$

but $e^{-2t}$ and $te^{-2t}$ are already accounted for in the characteristic solution. So we instead use

$y_p = (At^3+Bt^2)e^{-2t}$

which has derivatives

${y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}$

${y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}$

Substituting these into the left side of the ODE gives

$(4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}$

so that 6A = 12 and 2B = 0, or A = 2 and B = 0.

For the second solution corresponding to $-8t-12$ , we use

$y_p = Ct + D$

with derivative

${y_p}' = C$

${y_p}'' = 0$

Substituting these gives

$4C + 4(Ct+D) = 4Ct + 4C + 4D = -8t-12$

so that 4C = -8 and 4C + 4D = -12, or C = -2 and D = -1.

Then the general solution to the ODE is

$y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1$

Given the initial conditions y (0) = -2 and y' (0) = 1, we have

$-2 = C_1 - 1 \implies C_1 = -1$

$1 = -2C_1 + C_2 - 2 \implies C_2 = 1$

and so the particular solution satisfying these conditions is

$y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1$

or

$\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}$

7 0

2 years ago