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3 years ago

Why is it important to know whether a number is positive or negative

1 answer:

6 0

its good to know if a number is positive or negative because something as simple as forgetting the signs such as - + could drastically change your answer

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The Tigers’ second play went for +4 1/2 yards. Did the Tigers gain or lose yards on that play? How many yards did they gain or l

Vlad1618 [11]

Answer:

They gained 4 1/2 yards

Step-by-step explanation:

It was a positive gain because they didn't go back.

7 0

2 years ago

Read 2 more answers

Given f (x) = -5x + 1, find f (6)

Readme [11.4K]

Answer:

x=6(given)

-5x+1

-5*6+1

-30+1

-29

hope this answer helps u

pls mark it as brainliest

6 0

3 years ago

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How are the functions f(x)=16^x and g(x)=16^(1/2)x related? The output values of g(x) are one-half the output values of f(x) for

Goshia [24]

F(x) =16ˣ and g(x) = 16⁽ˣ/₂⁾

Since 16 = 2⁴, then we can write:

f(x) =2⁽⁴ˣ⁾ and g(x) = 2⁽⁴ˣ/₂⁾ = 2²ˣ

for x = 1 f(x) = 2⁴ = 16
for x = 1 g(x) = 2² = 4
(√16 = 4)

for x = 2 f(x) = 2⁸ = 256
for x = 2 g(x) = 2⁴ =16
(√256) = 16

for x = 3 f(x) = 2¹² = 4096
for x = 1 g(x) = 2⁶ = 64
(√4096 = 64)

We notice that:
The output values of g(x) are the square root of the output values of f(x) for the same value of x.

6 0

3 years ago

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(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

NIE

Vaselesa [24]

Hi!

Please see the attached image for your answer with explanation.

Have a great day!

4 0

2 years ago