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3 years ago

Evaluate 2^2 3^2 4^2

Mathematics

2 answers:

zlopas [31]3 years ago

4 0

Answer:29

Step-by-step explanation:

WARRIOR [948]3 years ago

4 0

2^2 = 4 3^2 = 9 4^2 = 16

Step-by-step explanation:

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Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

Recall that:

$y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

Then

$y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)$

$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

$\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

3 0

2 years ago

You are walking from home to a grocery store . You stop for a rest after 1/3 miles. The grocery store is actually 3/8 miles from

IgorLugansk [536]

Answer:

1/24

Step-by-step explanation:

1/3 = 8/24

3/8 = 9/24

9/24 - 8/24 = 1/24

3 0

2 years ago

A soccer game is 90 minutes long. 36 minutes have passed. What percentage of the game has passed?

Alex Ar [27]

Answer:

40%

Step-by-step explanation:

Percentage Passed = 36 ÷ 90 × 100 = 40%

6 0

3 years ago

How are a debit and credit represented on a number line?

const2013 [10]

Step-by-step explanation:

A credit is represented as a positive number that is located to the left of zero.

C) A credit is represented as a negative number that is located to the left of zero. A debit is represented as a positive number that is located to the left of zero.

3 0

3 years ago

dos aviones parten de una ciudad con direcciones S32°E y E57°N¿ cuál es el angulo que forma sus direcciones?

Elan Coil [88]

Answer:

El ángulo formado entre las dos direcciones es aproximadamente 115º.

Step-by-step explanation:

Las direcciones dadas en el enunciado significan lo siguiente:

S32ºE - <em>32º al este del sur.</em>

E57ºN - <em>57º al norte del este.</em>

Vectorialmente hablando, cada dirección es la siguiente:

S32ºE

$A(x,y) = (\sin 32^{\circ}, -\cos 32^{\circ})$

$A(x,y) = (0.530, -0.848)$

E57ºN

$B(x,y) = (\cos 57^{\circ},\sin 57^{\circ})$

$B(x,y) = (0.545, 0.839)$

El ángulo formado entre los dos vectores unitarios ( $\theta$ ), medido en grado sexagesimales, puede determinarse mediante la siguiente ecuación vectorial:

$\theta = \cos^{-1} \frac{A\,\bullet \,B}{\|A\|\cdot \|B\|}$

Donde:

$\|A\|$ - Norma de $A(x,y)$ .

$\|B\|$ - Norma de $B(x,y)$ .

Si sabemos que $A(x,y) = (0.530, -0.848)$ , $B(x,y) = (0.545, 0.839)$ , $\|A\| = 1$ y $\|B\| = 1$ , entonces el ángulo formado entre los dos vectores es:

$\theta = \cos^{-1} \frac{(0.530)\cdot (0.545)+(-0.848)\cdot (0.839)}{(1)\cdot (1)}$

$\theta \approx 115^{\circ}$

El ángulo formado entre las dos direcciones es aproximadamente 115º.

4 0

2 years ago