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2 years ago

Which mathematical property is demonstrated?

Mathematics

1 answer:

Arlecino [84]2 years ago

8 0

Answer:

I am pretty sure that the answer to this question is A: comunative property of addition

Step-by-step explanation:

I'm 99% sure

I'm sorry if it turns out to be wrong

cause I have not yet finished the test

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Graph the equation using the slope and the y intercept 2x-3y=8

sammy [17]

Step-by-step explanation:

<h2>Find slope-intercept structure </h2>

y=
$\frac{2}{3} x - \frac{8}{3}$
m=2/3, b=-8/3
therefore the slope will be 2/3

To find y-intercept

let x=0
2(0)-3y=8
-3y=8
y=-3/8

To find x-intercept

let y=0
2x-3(0)=8
2x=8
x=4

7 0

1 year ago

Two lines intersect at a point, forming ∠1 , ∠2 , ∠3 , and ∠4 . ∠1 and ∠3 are vertical angles. ∠2 and ∠4 are vertical

ZanzabumX [31]

Angle 1 and angle 2 are a linear pair, which make a sum of 180
angle 2=180-angle 1=180-70=110

6 0

2 years ago

Read 2 more answers

A function has y-intercept of -2 and a slope of -3/5. Which equation below describes the function?

alexandr402 [8]

D.y = -3/5x - 2

mggfffffffffffffffffffffffffffffffffffffffff

6 0

2 years ago

Solve y = x + 6 for x.

elena-s [515]

Answer:

Step-by-step explanation:

move the x over

-x+y=6

move the y over

-x = -y+6

times each number by a -

x=y-6

4 0

2 years ago

Read 2 more answers

(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

Aloiza [94]

Answer

(a) $F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

Step-by-step explanation:

(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 f(s)e^{-st} \, dt$

where a = ∞

=> $F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt$

where d(t) = δ(t)

=> $F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt$

Integrating, we have:

=> $F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.$

Inputting the boundary conditions t = a = ∞, t = 0:

$F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $f(t) = e^{-t} + 4e^{-4t} + te^{-3t}$

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt$

$F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt$

$F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt$

Integrating, we have:

$F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.$

Inputting the boundary condition, t = a = ∞, t = 0:

$F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

3 0

2 years ago