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FrozenT [24]
3 years ago
12

An expression is shown. 590.92 - 219.38 What is the value of the expression?​

Mathematics
2 answers:
ollegr [7]3 years ago
5 0

Answer:

371.54

Step-by-step explanation:

590.92 - 219.38 = 371.54

statuscvo [17]3 years ago
3 0

371.54 is the answer :)

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4 years ago
Graph the line with a slope of −1 that contains the point (3,6).
aksik [14]

Answer:

The equation of the line is y = -1.x + 9

Graph is provided in the attached figure

Step-by-step explanation:

The slope intercept equation of a line in 2D(x,y) coordinates is given by the equation

y = mx + c

where m is the slope of the line and c the y-intercept i.e. where the line crosses the y axis at x = 0

Given slope = -1, we can find c and the equation of the line

Since (3,6) is a point on the graph, these coordinates must satisfy the above equation

Substitute for y = 6 and x = 3

6 = (-1)3 + c\\\\c = 9\\\\\textrm{Equation of line is }\\y = -1.x + 9 \\y = 9-x

In the attached figure you can see that (3,6) is on the line

7 0
1 year ago
I need help on this question and every time I get help on this question some people get it wrong. can someone give me the right
LenaWriter [7]

Answer:

x = 8

y = -7

Step-by-step explanation:

  • (-3y) - 4x = -11

3y - 5x = -61

-----------------------+

-9x = -72

x = 8

  • 3y - 5x = -61

3y -5(8) = -61

3y - 40 = -61

3y = -21

y = -7

8 0
3 years ago
Which expression is equivalent to ^5 square root 13^3
joja [24]

Answer:

\sqrt[5]{13^3} = 13^{\frac{3}{5}}

Step-by-step explanation:

5 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:

<em>y''</em> + 4<em>y'</em> + 4<em>y</em> = 0

has characteristic equation

<em>r</em> ² + 4<em>r</em> + 4 = (<em>r</em> + 2)² = 0

with a double root at <em>r</em> = -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 <em>ansatz</em>

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 6<em>A</em> = 12 and 2<em>B</em> = 0, or <em>A</em> = 2 and <em>B</em> = 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 4<em>C</em> = -8 and 4<em>C</em> + 4<em>D</em> = -12, or <em>C</em> = -2 and <em>D</em> = -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 <em>y</em> (0) = -2 and <em>y'</em> (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
3 years ago
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