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3 years ago

Find the magnitude of WX for W(-2, 8, -3) and X(1, 4, -1)

Mathematics

2 answers:

zheka24 [161]3 years ago

8 0

The magnitude of WX is $\sqrt{29}$ ⇒ c

Step-by-step explanation:

The magnitude of XY (its length) where,

$X=(x_{1},y_{1},z_{1})$
$Y=(x_{2},y_{2},z_{2})$

is IXYI = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

∵ W = (-2 , 8 , -3)

∵ X = (1 , 4 , -1)

- To find the magnitude of WX use the formula above

∴ $x_{1}$ = -2 and $x_{2}$ = 1

∴ $y_{1}$ = 8 and $y_{2}$ = 4

∴ $z_{1}$ = -3 and $z_{2]$ = -1

- Substitute these values in the rule above

∵ IWXI = $\sqrt{(1--2)^{2}+(4-8)^{2}+(-1--3)^{2}}=\sqrt{9+16+4}$

∴ IWXI = $\sqrt{29}$

The magnitude of WX is $\sqrt{29}$

Learn more:

You can learn more about the position of an object in brainly.com/question/10940255

#LearnwithBrainly

mr_godi [17]3 years ago

8 0

Answer:

Step-by-step explanation:

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3 years ago

Find the Laplace transformation of each of the following functions. In each case, specify the values of s for which the integral

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Answer:

a. $\frac {2} {s-1}$ converges to s> 1.

b. $\frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $- \frac {2}{s + 3}$ converges to s> - 3.

d. $\frac {s}{s^2 + 25}$ converges to s> 0.

e. $\frac {10} {s^2 + 1}$ converges even s> 0.

f. $\frac {12}{s^2 + 4}$ converges to s> 0.

g. $-\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $\frac {1} {s ^ 2 + 4}$ converges to s> 0.

Step-by-step explanation:

a. $L \left\{2e^t \right\} = 2L \left\{e^t \right\} = 2 \cdot \frac {1} {s-1} = \frac {2} {s-1}$ converges to s> 1.

b. $L \left\{3e^{5t-3} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = \frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $L \left\{-2e^{-3t} \right\} = -2L \left\{e^{-3t} \right\} = - \frac {2}{s + 3}$ converges to s> - 3.

d. $L \left\{\cos\left (5t \right)\right\} = \frac {s}{s^2 + 25}$ converges to s> 0.

e. $L \left\{10 \sin\left(t\right)\right\} = 10L\left\{\sin\left(t\right)\right\} = \frac {10} {s^2 + 1}$ converges even s> 0.

f. $L \left\{6\sin \left(2t \right) \right\} = 6L\left\{\sin\left (2t\right)\right\} = \frac {12}{s^2 + 4}$ converges to s> 0.

g. $L \left\{-5\cos\left(2t + 1\right) \right\} = -5L\left\{\cos\left(2t + 1 \right)\right\} = -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $L\left\{\sin \left(t\right)\cos \left(t\right)\right\} = L\left\{\sin\left(2t\right)\frac{1}{2}\right\} =\frac{1}{2}\cdot \frac{2}{s^2+4} = \frac {1} {s ^ 2 + 4}$ converges to s> 0.