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3 years ago

Given the function f(x) = −5x2 − x + 20, find f(3).−28−136264

Mathematics

2 answers:

Flura [38]3 years ago

6 0

f(x) = −5x2 − x + 20, find f(3).

Substitute x = 3 into the function

f(3) = −5(3)^2 − 3 + 20

f(3) = -45 - 3 + 20

f(3) = -28

Answer

−28

Maslowich3 years ago

5 0

Answer: -28

Step-by-step explanation: The value of given function at x = 3 is -28

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Which ratio does not belong with 2/6<br> a) 1/3 b) 3/9 c)6/15 d) 12/36

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C)6/15

Step-by-step explanation:

it reduces to 2/5 no 1/3

12/36 --> 1/3

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What is the measure of angle E, in degrees?

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4 years ago

Use the Pythagorean Theorem to find x.

Anarel [89]

Answer:

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3 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

3 years ago

What is the solution to X and Y in these equations 3x + 9y=60 and -3x + y=30

Roman55 [17]

Answer:

x= -7 y=9

Step-by-step explanation:

3x +9y= 60 [1]

-3x +y=30 [2] rewrite this as the following: y=30+3x

With equation [2] you substitute it into equation [1]

so it will now be: 3x+ 9(30 +3x) =60

3x+270 +27x=60

Group the like terms

3x+27x=60-270

30x=-210

x=-7

Use this answer into the same equation

So: 3(-7)+9y=60

-21 +9y=60

9y=60+21

9y=81

y=9

8 0

4 years ago