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3 years ago

What is the area of this figure? 5 in 6 in 15 in 16 in 10 in 20 in

Mathematics

2 answers:

andrew-mc [135]3 years ago

6 0

Answer:

230 in.²

Step-by-step explanation:

20 · 10 = 200 in.²

5 · 6 = 30 in.²

200 in.² + 30 in.²

IrinaK [193]3 years ago

4 0

Answer:

230 sq in

Step-by-step explanation:

You can find entire area (20 x 16) and subtract out the area of the missing piece (15 x 6)

20 x 16 = 320

15 x 6 = 90

320 - 90 = 230

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3 years ago

6) Find b(10) in the sequence given by b(n) = -5 + 9(n-1).<br> b(10) =?

Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

1. Replace n with 10. The equation will look like:

-5+9(10-1)

2. Remember PEMDAS! The first step in solving the equation is solving whatever is in the parenthesis:

-5+9(9)

3. The next step is exponents, but since there aren't exponents in the equation, we move to multiplication and division. The equation should now look like this:

-5+81

4. Now we move on to addition and subtraction; add the two numbers!

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There's your answer!! Hope this helped :)

6 0

3 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:

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

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

3 years ago

What is the area of this figure? 5 in 6 in 15 in 16 in 10 in 20 in​

What is the area of this figure? 5 in 6 in 15 in 16 in 10 in 20 in