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4 years ago

Help on (b) pleasea level maths factor theorem

Mathematics

1 answer:

lorasvet [3.4K]4 years ago

6 0

Step-by-step explanation:

According to factor theorem, if x − c is a factor of f(x), then f(c) = 0.

Conversely, if f(c) = 0, then x −c is a factor of f(x).

a) If x − 3 is a factor of f(x), then f(3) will be 0.

f(3) = 4(3)³ − 12(3)² + 2(3) − 6

f(3) = 0

b) Having shown that x − 3 is a factor of f(x), we can use grouping or long division to factor f(x):

f(x) = 4x³ − 12x² + 2x − 6

f(x) = 4x² (x − 3) + 2 (x − 3)

f(x) = (4x² + 2) (x − 3)

Since 4x² + 2 has only imaginary zeros, 3 is the only real root.

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This box can be packed with 48 units cubes. The edge length of each unit cube is 1 meter. What is the volume of the box?

Dafna1 [17]

48 m³

Step-by-step explanation:

A cube has equal length width and height

A cube in this case has length 1m meaning the width and height are also 1 m each in length.

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If 48 of these can fit in the bigger box, then the bigger box has the volume of;

48 * 1 m³ = 48 m³

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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:

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

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3 years ago

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choli [55]

X=4 is your answer because 27 minus 7 is 20 and 20 divided by 5 is 4 have a good day

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Help on (b) pleasea level maths factor theorem ​

Help on (b) pleasea level maths factor theorem