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2 years ago

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Choose x = (x1, x2, x3, x4) in R4. It has 24 rearrangements like (x2, x1, x3, x4) and (x4, x3, x1, x2). Those 24 vectors, includ

ing x itself, span a subspace S. Find specific vectors x so that the dimension of S is: (a) 0, (b) 1, (c) 3, (d) 4.

1 answer:

zavuch27 [327]2 years ago

4 0

a. Take $\mathbf x$ to be the zero vector.

b. Fix one component $x_i$ from $i\in\{1,2,3,4\}$ , then set $x_j=0$ for $i\neq j$ . One such choice would be $\mathbf x=(0,x_2,0,0)$ .

c. Similar strategy as in (b), but we need 3 vectors in order for $S$ to have dimension of 3. One such choice would be $\mathbf x\in\{(x_1,0,0,0),(0,x_2,0,0),(0,0,0,x_4)\}$ .

d. You can use the example from (c); then you just need to insert the vector $(0,0,x_3,0)$ .

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answer A radio station located 120 miles due east of Collinsville has a listening radius of 100 miles. A straight road joins Col

melamori03 [73]

Answer:

168.7602 miles

Step-by-step explanation:

One way to solve this problem is by using an equation that describes the listening radius of the station, and another for the road, then the points where this two-equation intersect each other will represent when the driver starts and stops listening to the station, and the distance between the points is the miles that the driver will receive the signal.

The equation for the listening radius (the radio station is at (0,0)):

$x^2+y^2=100^2$

The equation for the road that past through the points (-120,0) and (80,100) (Collinsville and Harmony respectively):

$m=\frac{y_2-y_1}{x_2-x_1} =\frac{100-0}{80-(-120)}=\frac{100}{200}=\frac{1}{2}$

$y-y_1=m(x-x_1)\\y-0=\frac{1}{2}(x-(-120))\\ y=\frac{1}{2}x+60$

Substitutes the value of y in the equation of the circle:

$x^2+(\frac{1}{2}x+60)^2=100^2\\x^2+\frac{1}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600-10000=0\\\frac{5}{4} x^2+60x-6400=0\\5 x^2+240x-25600=0\\x^2+48x-5120=0\\$

The formula to solve second-degree equations:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a} \\x_{1,2}=\frac{-48\pm\sqrt{48^2-4(1)(-5120)} }{2(1)}\\x_{1,2}=\frac{-48\pm\sqrt{2304+20480} }{2}\\x_{1,2}=\frac{-48\pm\sqrt{22784} }{2}\\x_{1,2}=\frac{-48\pm16\sqrt{89} }{2}\\x_{1,2}=-24\pm8\sqrt{89} \\x_1=-24+8\sqrt{89}\approx51.4718\\x_2=-24-8\sqrt{89}\approx-99.4718\\$

Using the values in x to find the values in y:

$y_1=\frac{1}{2}x_1+60\\y_1=\frac{1}{2}(-24+8\sqrt{89} )+60\\y_1=-12+4\sqrt{89}+60\\ y_1=48+4\sqrt{89}\approx85.7359$

$y_2=\frac{1}{2}x_2+60\\y_2=\frac{1}{2}(-24-8\sqrt{89} )+60\\y_1=-12-4\sqrt{89}+60\\ y_1=48-4\sqrt{89}\approx10.2641$

The distance between the points (51.4718,85.7359) and (-99.4718,10.2641) :

$d=\sqrt{(x_1 -x_2 )^2+(y_1 -y_2)^2} \\d=\sqrt{(-24+8\sqrt{89} -(-24-8\sqrt{89}) )^2+(48+4\sqrt{89} -(48-4\sqrt{89}) )^2}\\d=\sqrt{(-24+8\sqrt{89} +24+8\sqrt{89} )^2+(48+4\sqrt{89} -48+4\sqrt{89} )^2}\\d=\sqrt{(16\sqrt{89} )^2+(8\sqrt{89} )^2}\\d=\sqrt{22784+5696}\\d=\sqrt{28480}\\d=8\sqrt{445}\approx168.7602miles$

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Ksenya-84 [330]

Tonya beat Jada by 5.13 seconds
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Answer:

The volume of platform

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Step-by-step explanation:

The height of the Platforms are

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Volume of the food prep area platform = Area × height = 5 b·d ft³

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Answer:

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2 years ago

M2/n2 is equivalent to _____.

tankabanditka [31]

Answer:

$\frac{m^2}{n^2}$ is equivalent to $m^2 \cdot n^{-2}$

Step-by-step explanation:

Using exponent rules:

$a^{-n}=\frac{1}{a^n}$

Given the expression:

$\frac{m^2}{n^2}$

⇒ $\frac{m^2}{n^2} = m^2 \cdot \frac{1}{n^2}$

Apply the exponent rule, we have;

$m^2 \cdot n^{-2}$

Therefore, the given expression $\frac{m^2}{n^2}$ is equivalent to $m^2 \cdot n^{-2}$

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2 years ago