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3 years ago

2(5x - 3) - 4 = x + 4 solve for x and show work

2 answers:

6 0

1. distribute --> 10x-6-4=x+4
2. do -6-4 --> 10x-10=x+4
3. add 10 on both sides --> 10x=x+4+10
4. subtract x on both sides and simplify equation --> 9x=14
5. divide both sides by 9 --> x=14/9

Katen [24]3 years ago

4 0

Answer:

x= 14/9 = 1 5/9

Step-by-step explanation:

Use the distributive property to multiply 2 by 5x−3.

2(5x−3)−4=x+4

Subtract 4 from −6 to get −10.

10x−6−4=x+4

Subtract x from both sides

10x−10=x+4

Combine 10x and −x to get 9x.

10x−10−x=4

Add 10 to both sides.

9x−10=4

Divide both sides by 9.

x= 14/9

Please mark brainliest.

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R-r=10

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2 years ago

BRAINLIEST GIVEN TO BEST ANSWER

lbvjy [14]

Answer:

Slope = $\frac{1}{4}$

Step-by-step explanation:

Slope of a straight line passing two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the formula,

m = $\frac{y_2-y_1}{x_2-x_1}$

From the graph attached,

Line is passing through two points (20, 5) and (80, 20),

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m = $\frac{20-5}{80-20}$

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The slope is $\frac{1}{4}$ .

3 0

2 years ago

(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:

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