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Akimi4 [234]
3 years ago
9

Pre-Cal Help!!! (See Attachment)

Mathematics
1 answer:
Verizon [17]3 years ago
7 0
In order to arrange the given vectors (u, v, w, x, y, and z) in ascending order of the magnitudes of their vector sums with vector t, you have to follow this procedure:

1) find the coordinates of vector t

2) find the coordinates of the other given vector (u, v, w, x, y, or z)

3) add the corresponding coordinates

4) find the magnitude of the sum vector using the formula

Let's do it:

1) vector t

x-coordinate = 4m/s * cos(60°) = 2 m/s
y-coordinate = 4m/s * sin(60°) = 3.46 m/s

2) u+t

vector u
: 3m/s, angle 120°=> second quadrant

x-coordinate = - 3 m/s * cos(60°) = -1.5 m/s
y-coordinate = 3m/s * sin(60°) = 2.6 m/s

u + t:
x-coordinate = 2 - 1.5 = 0.5
y-coordinate = 3.46 + 2.6  = 6.06

magnitude =  \sqrt{(0.5)^2+(6.06)^2} =6.08

3) v + t

vector v: 4.5 m/s , angle 135°

x-coordinate = - 4.5 * cos(180°-135°) = -4.5 * cos(45°) = -3.18
y-coordinate = 4.5 * sin (180° - 135°) = 4.5 * sin(45°) = 3.18

v + t
x-coordinate = 2 - 3.18 = - 1.18
y-coordinate = 3.46 + 3.18 = 6.64

magnitude= \sqrt{(-1.18)^2+(6.64)^2}= 6.74

4) w + t

vector w: 4m/s, angle 45°

x-coordinate = 4 * cos(45°) = 2.83
y-coordinate = 4 * sin(45°) = 2.83

vector w + t
x-coordinate = 2+2.83 = 4.83
y-coordinate = 3.46+2.83 = 6.29

magnitude= \sqrt{ (4.83)^{2} + (6.29)^{2} } =7.93

5) x + t

vector x: 6 m/s, angle 210°C => third quadrant

x-coordinate = - 6 * cos(30°) = - 5.2
y-coordinate = - 6 * sin(30°) = - 3

x+t
x-coordinate = 2 - 5.2 = - 3.2
y-coordinate = 3.16 - 3  = - 0.16

magnitude= \sqrt{ (-3.2)^{2} + (-0.16)^{2} } =3.20

6) y+t

vector y: 5m/s, angle 330°=> fourth quadrant

x-coordinate = 5 *cos(360° - 330°) = 4.33
y-coordinate = - 5 * sin(30°) = -2.5

vector y+t
x-coordinate = 4.33 + 2 = 6.33
y-coordinate = - 2.5 + 3.46 = 0.96

magnitude= \sqrt{ 6.33^{2} + 0.96^{2} } =6.40

7) z+t

vector z = 7m/s angle 240° => third quadrant

x-coordinate = - 7 * cos(60°) = - 3.5
y-coordinate = - 7 * sin(60°) = - 6.06

vector z + t
x-coordinate = 2 - 3.5 = - 1.5
y-coordinate = 3.46 - 6.06 = -2.6

magnitude = 3.00

Now, you have all the numbers and just have to order them.

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The values of a, b and c are:

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Step-by-step explanation:

Step 1:

First Confirm the sequence is quadratic by finding the second difference.

Sequence = 1,11,27,49

1st difference:

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2nd difference:

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  • 22-16=6

Step 2:

Just divide the second difference by 2, you will get the value of a.

6 ÷ 2 = 3

So the first term of the nth term is 3n^{2}

Step 3: Next, substitute the number 1 to 5 into  3n^{2}

n = 1,2,3,4,5

3n² = 3,12,27,48,75

Step 4:

Now, take these values (3n²) from the numbers in the original number sequence and work out the nth term of these numbers that form a linear sequence.

n = 1,2,3,4,5

3n² = 3,12,27,48,75

Differences:

1 - 3 = -2

11 - 12 = -1

27 - 27 = 0

49 - 48 = 1

Now the nth term of these differences (-2,-1,0,1) is (n - 1) - 2.

so b = 1, and c = -3

Step 5: Write down your final answer in the form an² + bn + c.

3n² + (n - 1) -2

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Therefore, the values of a, b and c are:

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y=z=0\implies x=2\implies (2,0,0)

x=z=0\implies y=2\implies(0,2,0)

x=y=0\implies 2z=2\implies z=1\implies(0,0,1)

Plane 2, 4<em>x</em> + 4<em>y</em> + <em>z</em> = 8:

y=z=0\implies4x=8\implies x=2\implies(2,0,0)

x=z=0\implies4y=8\impliesy=2\implies(0,2,0)

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Both planes share the same <em>x</em>- and <em>y</em>-intercepts, but the second plane's <em>z</em>-intercept is higher, so Plane 2 acts as the roof of the bounded region.

Meanwhile, in the (<em>x</em>, <em>y</em>)-plane where <em>z</em> = 0, we see the bounded region projects down to the triangle in the first quadrant with legs <em>x</em> = 0, <em>y</em> = 0, and <em>x</em> + <em>y</em> = 2, or <em>y</em> = 2 - <em>x</em>.

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=\displaystyle\int_0^2\left(7-7x+\frac74 x^2\right)\,\mathrm dx=\boxed{\frac{14}3}

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